Does $\exp(\ln(I+A))=I+A$ when $\|A\|<1$? For matrices, I know certain equalities like $e^{A+B}=e^Ae^B$ aren't always true. I'm curious, do $\exp$ and $\ln$ serve as inverses?
I saw earlier that if $\|A\|<1$, then $\ln(I+A)$ converges. My question is, does
$$
\exp(\ln(I+A))=I+A
$$
when $\|A\|<1$? Certainly $\exp(\ln(I+A))$ converge since $\ln(I+A)$ converges and $\exp(\cdot)$ converges for all matrices, but do we necessarily get the original matrix back?
 A: Yes, you do get the original matrix back. One way to see this is to recall that it is true for the reals and these are analytic functions (the one with log suitably restricted). 
So you can develop both $\exp(x)$ and $\ln(1+x)$ into powerseries (the former converges everywhere and the latter for $|x|<1$). 
But now you could consider these powerseries as formal powerseries, say denoting them by $E(X)$ and $L(X)$, respectively. 
Then you can check (or rather it also follows) from the result for the reals that $E(L(X)) =1+X$ as formal powerseries. 
So it also holds when you plug in a matrix (such that the series converge).
Note: you recalled the exponential identity that does not always hold, and one might now think my argument is bogus as it would also show the former, but it is not as it does not show it, since  while it is true that $\exp(X+Y)=\exp(X)\exp(Y)$ as formal powerseries the variablex $X,Y$ are assumed to commute. So, you need this too for the matrices. And if it is true that $AB=BA$ then infact $\exp(A+B)=\exp(A)\exp(B)$.
A: Another more analytical approach than quid's fine answer is the following:
In the unit ball of $\text{Mat}_n(\mathbb{C})$ the diagonalizable matrices (that are in the ball) are dense. It is trivial to verify the equation 
$$\exp(\log(1+A))=1+A$$
holds for $A$ in the unit ball, and diagonalizable. Then, you can use the fact that both $X\mapsto \exp(\log(1+X))$ and $X\mapsto 1+X$ are continuos functions from the unit ball to $\text{Mat}_n(\mathbb{C})$. Since the latter space is metrizable (Hausdorff) and the diagonalizable matrices are dense, it follows that $X\mapsto \exp(\log(1+X))$ and $X\mapsto 1+X$ must agree on the unit ball.
